Siemens SIMOTICS L-1FN3 Configuration Manual page 128

Configuration
5.3 Examples
Calculating the required maximum force for the selected primary sections
1st iteration step
Peak load motor
1FN3100-4WC00-0BA1
Continuous load motor
1FN3150-3NC70-0BA1
The force of the primary sections previously selected is too low, both for the peak load motor
and the continuous load motor. Therefore, a new primary section has to be selected.
2nd iteration step
New, improved motor selection for the example (see motor data sheets):
Peak load motor
Continuous load motor 1FN3150-4NB80-0BA1
Peak load motor
1FN3100-5WC00-0BA1
1FN3150-4WC00-0BA1
Continuous load motor
1FN3150-4NB80-0BA1
The further calculations in this example are performed with the peak load motor 1FN3150-
4WC00-0BA1 or the continuous load motor 1FN3150-4NB80-0BA1.
126
The total mass to be moved m
m = m + m = (50 + 8.5) kg = 58.5 kg
ges Motor
The maximum force that the motor must supply for the duty cycle is:
F = m · a + F = 58.5 kg · 41 m/s
L,MAX ges r
F = 2499 N
L,MAX
The total mass to be moved m
m = m + m = (50 + 11.7) kg = 61.7 kg
ges Motor
The maximum force that the motor must supply for the duty cycle is:
F = m · a + F = 61.7 kg · 41 m/s
L,MAX ges r
F = 2630 N
L,MAX
Article No.
1FN3100-5WC00-0BA1
1FN3150-4WC00-0BA1
m = 60.4 kg
ges
F = 2576 N (no control reserve)
L,MAX
m = 61.4 kg
ges
F = 2617 N (10% control reserve present)
L,MAX
(calculation uses same approach as in the 1st iterative step)
m = 65.3 kg
ges
F = 2777 N
L,MAX
(calculation uses same approach as in the 1st iterative step)
:
ges
+ 100 N
2
:
ges
+ 100 N
2
v F
MAX, FMAX MAX
109 m/min 2750 N
126 m/min 3300 N
109 m/min 3060 N
Configuration Manual, 10/2018, 6SN1197-0AB86-0BP2
m
Motor
(with precision cooler)
10.4 kg
11.4 kg
15.3 kg
1FN3 linear motors